What is the general solution for cos^2(x)=((4k-3)/9) ?

The general solution for cos^2(x)=((4k-3)/9) is

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cos^2(x)=((4k-3)/9)

cos^{2} (x) = (\frac{4 k - 3}{9})

x = arccos (\frac{4 k - 3}{3}) + 2 πn, x = - arccos (\frac{4 k - 3}{3}) + 2 πn, x = arccos (- \frac{4 k - 3}{3}) + 2 πn, x = - arccos (- \frac{4 k - 3}{3}) + 2 πn

Solution steps

cos^{2} (x) = (\frac{4 k - 3}{9})

Solve by substitution

cos (x) = \frac{4 k - 3}{3}, cos (x) = - \frac{4 k - 3}{3}

cos (x) = \frac{4 k - 3}{3} : x = arccos (\frac{4 k - 3}{3}) + 2 πn, x = - arccos (\frac{4 k - 3}{3}) + 2 πn

cos (x) = - \frac{4 k - 3}{3} : x = arccos (- \frac{4 k - 3}{3}) + 2 πn, x = - arccos (- \frac{4 k - 3}{3}) + 2 πn

Combine all the solutions

x = arccos (\frac{4 k - 3}{3}) + 2 πn, x = - arccos (\frac{4 k - 3}{3}) + 2 πn, x = arccos (- \frac{4 k - 3}{3}) + 2 πn, x = - arccos (- \frac{4 k - 3}{3}) + 2 πn

9 = - 9 \cdot sin (2 x)

2 sin (4 t) = 2 sin (4 t) cos (4 t)

a^{1} sin (x) = 43.1253

52800 = \frac{( 1500 ) ^{2} sin ( x )}{32.2}

7 cos (3 x) = 2, 0 \leq x < 2 π

What is the general solution for cos^2(x)=((4k-3)/9) ?
The general solution for cos^2(x)=((4k-3)/9) is